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4 Improved text.

I think you can get a much faster (and maybe easier...) proof using Riemannian geometry, as follows:

First, recall that a semi-simple connected Lie group $G$ is compact if and only if its Killing form $B$ is negative-definite (the proof is easy, see, e.g., Thm 2.28 in these notes). The implication you need side we will use ($G$ compact semi-simple $\Rightarrow$ $B$ neg.-def.) actually follows directly from $B(X,X)=tr(ad(X)\cdot ad(X))$ using an orthonormal basis with respect to an auxiliary bi-invariant metric to compute this trace.

The

Now, the Ricci curvature of any bi-invariant metric on $G$ (that exists because $G$ is compact) can be computed as: $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same notes. By the first observation above, since $G$ is compact and semi-simple, its Killing form $B$ is negative-definite. Hence the above formula gives $Ric>0$. So, by the Bonnet-Myers Theorem, $G$ must have finite fundamental group. Q.E.D.

Perhaps this was the sketch of proof originally suggested to the OP?

3 added 235 characters in body

I think you can get a much faster (and maybe easier...) proof using basic Riemannian geometry(maybe this was the sketch of proof originally suggested to the OP?) , as follows:

First, recall that a semi-simple connected Lie group $G$ is compact if and only if its Killing form $B$ is negative-definite (the proof is quite simpleeasy, see, e.g., Thm 2.28 in these notes).

Since The implication you start with a need ($G$ compact semi-simple Lie group $G$, by compactness, it has a \RightarrowB$neg.-def.) actually follows directly from$B(X,X)=tr(ad(X)\cdot ad(X))$using an orthonormal basis with respect to an auxiliary bi-invariant metric to compute this trace. The Ricci curvature of this any bi-invariant metric on$G$(that exists because$G$is compact) can be computed as: $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same notes. By the first observation above, since$G$is compact and semi-simple, its Killing form$B$is negative-definite. Hence the above formula gives$Ric>0$. FinallySo, by the Bonnet-Myers Theorem,$G$must have finite fundamental group. Q.E.D. Perhaps this was the sketch of proof originally suggested to the OP? 2 Improved text I think you can do get a much faster proof using basic Riemannian geometry (maybe this was the sketch of proof originally suggested to the OP?) as follows. Let$G$be : First, recall that a semi-simple connected Lie group . Then$G$is compact if and only if its Killing form$B$is negative-definite (see, the proof is quite simple, see, e.g., Thm 2.28 in these notes). Since$G$is you start with a compact semi-simple Lie group$G$, by compactness, it has a bi-invariant metric, and the . The Ricci curvature of this metric is can be computed as:$Ric(X,Y)=-\frac14 B(X,Y)$($Ric(X,Y)=-\frac14 B(X,Y), see Remark 2.27 in the same notes)notes. Since By the first observation, since $G$ is compact, its Killing form $B$ is negative-definite, it follows that . Hence the above formula gives $Ric>0$, so Ric>0$. Finally, by the Bonnet-Myers Theorem,$G\$ must have finite fundamental group.

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